When photons of energy 4.25 eV strike the surface of metal, A, the ejected photoelectrons have maximum kinetic energy $T_{A}$ and de Broglie wavelength $λ_{A}$ .The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is $T_{B} = T_{A} - 1.50 e V$ . If the de Broglie wavelength of these photoelectron is $λ_{B} = 2 λ_{A},$ then

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the work function of A is 2.25 eV

the work function of B is 4.20 eV

$T_{A} = 2.00 e V$

$T_{B} = 2.75 e V$

answer is A, B, C.

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Detailed Solution

The threshold wavelength is 5200 $\overset{0}{A}$ . For ejection of electrons, the wavelength of the light should be less than 5200 $\overset{0}{A}$ so that frequency increases and hence the energy of incident photon increases. UV light has less
wavelength than $\overset{0}{A}$ .

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